How to solve $\int\tfrac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx$?

Question

I am trying to find a solution for the following integration for a physics problem. Could anyone give hint on how to do this? $$ \int\frac{\sqrt{x^2+1}}{\left(x^2+a\right)^\frac{3}{2}}dx $$

Thanks for your time.

Answer 1

Allow us to use generalized binomial expansion theorem on the $\sqrt{x^2+1}$ and the $u$ substitution $x^2+a=u$.

$$\sqrt{x^2+1}=\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}=x+\frac12x^{-1}-\frac18x^{-2}+\dots$$

$$\begin{align} \int\frac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx & = \int\frac{\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}}{(x^2+a)^{3/2}}dx \\ & = \frac12\sum_{n=1}^\infty\binom{1/2}{n}\int\frac{x^{-2n}}{(x^2+a)^{3/2}}(2xdx) \\ & = \frac12\sum_{n=1}^\infty\binom{1/2}{n}\int u^{-3/2}(u-a)^{-n}du \\ \end{align}$$

Again, binomial expand the $(u-a)^{-n}$

$$\begin{align} & = \frac12\sum_{n=1}^\infty\sum_{k=0}^\infty\binom{1/2}{n}\binom nk(-a)^k\int u^{-3/2}u^{-1-k}du \\ & = C+\frac12\sum_{n=1}^\infty\sum_{k=0}^\infty\binom{1/2}{n}\binom nk(-a)^k\frac{(x^2+a)^{-(3/2)-k}}{-(3/2)-k} \\ \end{align}$$

I guess you could say its no longer an integral problem and more of a summation problem.

Also, don't expect anything simple, because according to Wolframalpha, the integral involves elliptic integrals.